SOLUTION: The volume of the rectangular block in the figure is calculated from the following measurements of its dimensions: �10.00±0.10 cm, �5.00±0.06 cm and �4.00±

Question 1181807: The volume of the rectangular block in the figure is calculated from the following measurements of its dimensions: �10.00±0.10 cm, �5.00±0.06 cm and �4.00±0.04 cm. Calculate the error in the value of the volume of the block assuming: a) the errors are independent b) the errors are correlated such that they all push the estimate in the same direction.
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Here's how to calculate the error in the volume, considering both independent and correlated errors:
**1. Calculate the Nominal Volume:**
V = lwh = (10.00 cm)(5.00 cm)(4.00 cm) = 200.00 cm³
**2. Fractional Uncertainties:**
* Δl/l = 0.10 cm / 10.00 cm = 0.01
* Δw/w = 0.06 cm / 5.00 cm = 0.012
* Δh/h = 0.04 cm / 4.00 cm = 0.01
**a) Independent Errors:**
When errors are independent, we add the *squares* of the fractional uncertainties and then take the square root.
1. Fractional uncertainty in volume:
(ΔV/V)² = (Δl/l)² + (Δw/w)² + (Δh/h)²
(ΔV/V)² = (0.01)² + (0.012)² + (0.01)²
(ΔV/V)² = 0.0001 + 0.000144 + 0.0001 = 0.000344
ΔV/V = √0.000344 ≈ 0.0185
2. Absolute uncertainty in volume:
ΔV = (ΔV/V) * V = 0.0185 * 200.00 cm³ ≈ 3.7 cm³
3. Volume with uncertainty:
V = 200.0 ± 3.7 cm³
**b) Correlated Errors:**
When errors are fully correlated (all pushing the estimate in the same direction), we simply *add* the fractional uncertainties.
1. Fractional uncertainty in volume:
ΔV/V = Δl/l + Δw/w + Δh/h
ΔV/V = 0.01 + 0.012 + 0.01 = 0.022
2. Absolute uncertainty in volume:
ΔV = (ΔV/V) * V = 0.022 * 200.00 cm³ = 4.4 cm³
3. Volume with uncertainty:
V = 200.0 ± 4.4 cm³
**Summary:**
* **Independent Errors:** V = 200.0 ± 3.7 cm³
* **Correlated Errors:** V = 200.0 ± 4.4 cm³
As expected, the uncertainty is larger when the errors are correlated because they all contribute to the error in the same direction. When independent, there is a chance that some errors cancel each other out, thus reducing the total uncertainty.

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